Solution: Bernoulli Sums and the de Moivre-Laplace Approximation

Exercise: Bernoulli Sums and the de Moivre-Laplace Approximation

Part 1

For i.i.d. Bernoulli( $p$ ) with $p = 0.55$ :

\mathbb{E}[S_n] = np = 55, \qquad \operatorname{Var}(S_n) = np(1-p) = 100\cdot 0.55\cdot 0.45 = 24.75.

So $S_{100} \overset{d}{\approx} \mathcal{N}(55,\,24.75)$ , with standard deviation $\sqrt{24.75} \approx 4.975$ .

Without continuity correction:

\mathbb{P}(S_{100} \ge 60) \approx 1 - \Phi\!\left(\frac{60 - 55}{4.975}\right) = 1 - \Phi(1.005) \approx 1 - 0.8426 = 0.1574.

With continuity correction:

\mathbb{P}(S_{100} \ge 60) \approx 1 - \Phi\!\left(\frac{59.5 - 55}{4.975}\right) = 1 - \Phi(0.904) \approx 1 - 0.8170 = 0.1830.

from scipy.stats import binom

exact = 1 - binom.cdf(59, n=100, p=0.55)
print(f"Exact binomial: {exact:.4f}")
# Exact binomial: 0.1827

Comparison: exact

0.1827

, corrected Gaussian

0.1830

(off by

0.0003

), uncorrected Gaussian

0.1574

(off by

0.025

). The continuity correction is dramatically better — an order of magnitude more accurate for the same computational cost.

The CLT is a continuous approximation to a discrete distribution. Shifting by half a unit to straddle the integer (continuity correction) recovers almost all the accuracy lost to the discretisation.
At $n = 100$ the uncorrected approximation is already off by 15% relative. For smaller $n$ the error explodes; use exact binomial, not Gaussian, for $n \lesssim 25$ .
Always use the continuity correction on discrete CLT problems. The cost is typing ±0.5; the gain is an order of magnitude in accuracy.
This is the de Moivre-Laplace theorem — historically the first CLT, proved by de Moivre for fair coins ( $p = 1/2$ ) in 1733 and generalised by Laplace to arbitrary $p \in (0, 1)$ in 1810. It predates the general CLT by over a century.